Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to understand intuitively the notion of Lipschitz function.

I can't understand why bounded function doesn't imply Lipschitz function.
enter image description here

I need a counterexample or an intuitive idea to clarify my notion of Lipschitz function.

I need help

Thanks a lot

share|improve this question
also assume continuity? –  Halil Duru Apr 6 '13 at 20:38

5 Answers 5

A Lipschitz function is such that $$|f(x)-f(y)|\leq \alpha |x-y|$$ for any points you pick. Writing this as $$\left|\frac{f(x)-f(y)}{x-y} \right|\leq \alpha $$

what we're saying is that the slope of the secant line joining $(x,f(x))$ and $(y,f(y))$ is always bounded above by $\alpha$. An example of a Lipschitz function is $\sin x$, or $x$. An example of a function which is not Lipschitz but is bounded is $$\sin (x^2)$$ over $\Bbb R$. This is because as we go further towards $+\infty$, the oscillation becomes faster, and thus the slope of the secant lines get nearer and nearer to vertical ones.

An example of a function that is not Lipschitz nor bounded is $\sqrt x$ over $\Bbb R_{>0}$. This is because if we fix $x=0$ and make $y$ very close to $0$, the slope of the secant line grows without bound.

Finally, we can give an example of a function which is Lipschitz but isn't bounded: $x+\sin x$ over $\Bbb R$. Its slope will never get larger than $1$.

share|improve this answer
Excellent explanation! +1 –  Clayton Apr 6 '13 at 20:33
If a function has a derivative that is bounded, must it be a Lipschitz function? –  Peter Olson Apr 7 '13 at 0:28
@PeterOlson Assume $f$ has bounded derivative. By the Mean Value Theorem, for each $x,y\in\operatorname{dom}f$, what can you equate $$\frac{f(x)-f(y)}{x-y}$$ to? –  Pedro Tamaroff Apr 7 '13 at 0:34

Look at the square root function on $[0,1]$ and its behaviour at 0.

share|improve this answer

Look at the unitary circle, $f(x)=\sqrt{1-x^2}$ how is the tangent line in $x=1$?

share|improve this answer

Hint: $f$ differentiable and Lipschitz $\Rightarrow$ $f'$ bounded.

But $f$ bounded and differentiable $\not\Rightarrow$ $f'$ bounded.

share|improve this answer

With this version (just assuming boundedness) , any discontinuous bounded function forms a counter-example .[why?]

But if you also assume continuity , see the examples below/above.

Furthermore , you may even have differentiability and boundedness without Lipschitzness......

share|improve this answer
How could a function be differentiable without being lipschitz? –  treble Apr 6 '13 at 21:01
Look at Tamaroff's example $sin[x^2]$ ..Isn't it differentiable? –  Halil Duru Apr 6 '13 at 21:19
OK, I wasn't thinking about the derivative going to infinity. –  treble Apr 6 '13 at 22:28
Exactly.......... –  Halil Duru Apr 6 '13 at 22:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.